Rows are periods, columns are groups
Answer: it would be 0.026 moles
Explanation: PV=nRT, P is the pressure of gas, V is the volume it occupies n is the number of moles of gas present in the sample, R is the universal gas constant which is equal to 0.0821 atm L/mol K and T is the absolute temperature of the gas
Molar mass RbMnO₄ = 204.40 g/mol
1 mole ---------- 204.40 g
7.88 mole ------ ?
mass = 7.88 * 204.40 / 1
mass = 1610.672 g
hope this helps!
Answer:
Pp O2 = 82.944 KPa
Explanation:
heliox tank:
∴ %wt He = 32%
∴ %wt O2 = 68%
∴ Pt = 395 KPa
⇒ Pp O2 = ?
assuming a mix of ideal gases at the temperature and volumen of the mix:
∴ Pi = RTni/V
∴ Pt = RTnt/V
⇒ Pi/Pt = ni/nt = Xi
⇒ Pi = (Xi)*(Pt)
∴ Xi: molar fraction (ni/nt)
⇒ 0.68 = mass O2/mass mix
assuming mass mix = 100 g
⇒ mass O2 = 68 g
∴ molar mass O2 = 32 g/mol
⇒ moles O2 = (68 g)(mol/32 g) = 2.125 mol O2
⇒ mass He = 32 g
∴ molar mass He = 4.0026 g/mol
⇒ moles He = (32 g)(mol/4.0026 g) = 7.995 mol He
⇒ nt = nO2 + nHe = 2.125 mol + 7.995 mol = 10.12 moles
molar fraction O2:
⇒ X O2 = nO2/nt = (2.125 mol/10.12 mol) = 0.2099
⇒ Pp O2 = (X O2)(Pt)
⇒ Pp O2 = (0.2099)(395 KPa)
⇒ Pp O2 = 82.944 KPa
Answer:
The answer to your question is all the formulas in bold has the same empirical formula
Explanation:
Data
Empirical formula CH₂O
Process
To solve this problem factor the subscripts of each formula and compare the result with the empirical formula given.
a) C₂H₄O₂ factor 2 2(CH₂O)
b) C₃H₆O₃ factor 3 3(CH₂O)
c) CH₂O₂ this formula can not be simplified
d) C₅H₁₀O₅ factor 5 5(CH₂O)
e) C₆H₁₂O₆ factor 6 6(CH₂O)
